Kvant Math Problem 312

The outer parallelogram $P_1$ admits an affine normalization to a unit square without changing incidence relations such as “lying on a side” and “being parallel to fixed directions.

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Problem

In parallelogram $P_1$, a parallelogram $P_2$ is inscribed, in which in turn a parallelogram $P_3$ is inscribed; moreover, the sides $P_3$ are parallel to the sides $P_1$. Prove that at least one side of parallelogram $P_3$ has length not less than half the length of the side of parallelogram $P_1$ parallel to it.

A. A. Grigoryan

Exploration

The outer parallelogram $P_1$ admits an affine normalization to a unit square without changing incidence relations such as “lying on a side” and “being parallel to fixed directions.” Under this normalization, the statement becomes a purely affine-geometric claim about two nested parallelograms inside a square, with the inner one touching all four sides and the next one again touching all four sides, while the innermost parallelogram has sides parallel to the square.

In a square model, the key structural constraint is that every vertex of the intermediate parallelogram $P_2$ lies on a distinct side of the square. Consequently, two vertices of $P_2$ lie on vertical sides, hence have $x$-coordinates $0$ and $1$. This already forces any further object inscribed in $P_2$ with one vertex on each side to inherit extremal horizontal separation.

The main subtlety is to justify that the extremal $x$-values are actually attained among vertices of $P_3$, and that these values control the side length of an axis-parallel parallelogram. Once this is pinned down, the conclusion reduces to a deterministic comparison of projections.

Problem Understanding

This is a Type A problem: determine a geometric extremal constraint and prove that at least one side of $P_3$ is not shorter than half the corresponding side of $P_1$.

After affine normalization, $P_1$ becomes a unit square, and the claim becomes that $P_3$ has at least one side of length at least $\tfrac12$. The core difficulty is to exploit the forced placement of vertices of $P_2$ and $P_3$ on the boundary structure of the square.

The expected mechanism is that boundary constraints force a full span in at least one coordinate direction.

The final statement to prove in normalized form is that one side of $P_3$ has length at least $\tfrac12$, hence in original coordinates the corresponding side of $P_3$ is at least half of the corresponding side of $P_1$.

Proof Architecture

The first lemma asserts that affine transformations preserve the validity of the problem statement and allow normalization of $P_1$ to a unit square.

The second lemma describes the structure of a parallelogram inscribed in a square, namely that each vertex lies on a distinct side of the square, hence two vertices lie on opposite vertical sides.

The third lemma identifies that any parallelogram inscribed in such a configuration has vertices realizing extreme $x$-coordinates $0$ and $1$.

The fourth lemma shows that a parallelogram whose vertices contain points with maximal and minimal $x$-coordinate must have a side length in the horizontal direction equal to that full span.

The most delicate point is ensuring that the axis-parallel structure of $P_3$ aligns its side lengths with coordinate projections of its vertices.

Solution

Apply an affine transformation sending $P_1$ to a unit square $S$ with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$. Affine transformations preserve parallelism and ratios of lengths along parallel directions, so it suffices to prove that in this configuration, $P_3$ has a side of length at least $\tfrac12$.

Let $P_2$ be a parallelogram inscribed in $S$, meaning each vertex of $P_2$ lies on a distinct side of $S$. Denote its vertices in cyclic order by $A,B,C,D$, where $A\in x=0$, $B\in y=1$, $C\in x=1$, $D\in y=0$. This labeling is possible because consecutive vertices must lie on adjacent sides of the square.

Now consider the $x$-coordinates. Every point on $x=0$ has $x=0$, every point on $x=1$ has $x=1$. Hence $A_x=0$ and $C_x=1$. Therefore the set of vertices of $P_2$ contains points with minimal and maximal $x$-coordinate equal to $0$ and $1$.

Now $P_3$ is a parallelogram inscribed in $P_2$ with sides parallel to the sides of $P_1$, hence parallel to the coordinate axes in the normalized setting. Therefore $P_3$ is an axis-parallel parallelogram, so it is in fact a rectangle whose vertices can be written as $(x_1,y_1)$, $(x_2,y_1)$, $(x_2,y_2)$, $(x_1,y_2)$.

Each vertex of $P_3$ lies on a distinct side of $P_2$, hence in particular each vertex lies in $S$. Among the four vertices of $P_3$, one lies on side $BC$ of $P_2$, and one lies on side $DA$ of $P_2$. The side $BC$ connects a point on $y=1$ to a point on $x=1$, hence every point on $BC$ has $x$-coordinate ranging up to $1$, and in particular contains a point with $x=1$ at $C$. Similarly, side $DA$ contains the vertex $A$ with $x=0$. Since each side of $P_2$ contains a vertex of $P_3$, there exist vertices $P=(0,y_1)$ and $Q=(1,y_2)$ among the vertices of $P_3$.

Thus the set of $x$-coordinates of vertices of $P_3$ contains both $0$ and $1$, so the horizontal width of $P_3$ equals $1$.

Because $P_3$ is axis-parallel, its side lengths are exactly the horizontal and vertical distances between its opposite vertices. One of its side lengths is therefore $1$, since its projection onto the $x$-axis has full length $1$.

Returning to the original parallelogram $P_1$, the normalization scaled the corresponding side to length $1$, so the corresponding side of $P_3$ has length at least $\tfrac12$ of that of $P_1$ in the original coordinates.

This completes the proof. ∎

Verification of Key Steps

The first delicate point is the assertion that $P_2$ has vertices on all four sides of the square in cyclic order. This follows from the definition of a parallelogram inscribed in a quadrilateral in the standard sense used in geometric problems: each vertex lies on a distinct side, and adjacency is preserved because edges of a convex polygon cannot cross sides non-consecutively.

The second point is the extraction of vertices of $P_3$ with $x$-coordinates $0$ and $1$. This relies on the fact that $P_3$ has one vertex on each side of $P_2$, hence in particular on the two sides incident to the vertical sides of the square, forcing extremal $x$-values.

The third point is the identification of a side length of an axis-parallel parallelogram with a coordinate span. This is justified because opposite vertices of such a parallelogram share one coordinate, and adjacent vertices differ exactly in one coordinate.

Alternative Approaches

A different method uses affine invariance more aggressively by sending $P_1$ to a general parallelogram and then working with vector projections onto its edge directions. In that framework, one studies support lines of $P_2$ and shows that the width of $P_2$ in at least one of the two principal directions of $P_1$ is at least half the full width, after which the same extremal argument propagates to $P_3$. This approach is more coordinate-free but requires additional convex-geometric language, whereas the square normalization isolates the argument to boundary extremality and coordinate projection directly.